Let $(X,x_0)$ be a pointed simplicial set. Assume if you like that $X$ is the nerve of a category but do not assume that $X$ is a Kan complex.

Because $Ex^\infty X$ is a Kan complex, every homotopy class $\alpha \in \pi_n(X,x_0)$ may be represented by a map $sd^k \Delta[n] \to X$ such that the restriction $sd^k \partial \Delta[n] \to X$ is constant at $x_0$. I'm wondering about different "normal forms" for homotopy classes.

For instance, consider subidivided cubes. In dimension 2, I think they should look like this:

$\require{AMScd} D^2_0 = \begin{CD} \bullet \end{CD} \\ D^2_1 = \begin{CD} \bullet @>>> \bullet @<<< \bullet \\ @VVV @VVV @VVV \\ \bullet @>>> \bullet @<<< \bullet \\ @AAA @AAA @AAA \\ \bullet @>>> \bullet @<<< \bullet \\ \end{CD} \\ D^2_2 = \begin{CD} \bullet @>>> \bullet @<<< \bullet @>>> \bullet @<<< \bullet\\ @VVV @VVV @VVV @VVV @VVV \\ \bullet @>>> \bullet @<<< \bullet @>>> \bullet @<<< \bullet\\ @AAA @AAA @AAA @AAA @AAA \\ \bullet @>>> \bullet @<<< \bullet @>>> \bullet @<<< \bullet\\ @VVV @VVV @VVV @VVV @VVV \\ \bullet @>>> \bullet @<<< \bullet @>>> \bullet @<<< \bullet\\ @AAA @AAA @AAA @AAA @AAA \\ \bullet @>>> \bullet @<<< \bullet @>>> \bullet @<<< \bullet \end{CD} \\ D^2_3 = \dots $


  1. Can every $\alpha \in \pi_n(X, x_0)$ be represented by a map $D^n_k \to X$ sending the boundary to the constant at $x_0$?
  2. If not, is there a better definition of subdivided cubes for which the answer to (1) becomes "yes"?

It's nice that with the above definition, $D^n_{k+1}$ can be obtained by gluing together a bunch of copies of $D^n_k$ in an easy way. But perhaps this is too good to be true.

A trivial observation is that for $n=1$ the above cubical subdivision is basically the same as barycentric subdivision and the answer to (1) comes out as "yes".

  • $\begingroup$ Why not just convert X to a cubical set (applying the cubical nerve functor), represent α as a subdivided cube there, then go back using the realization functor? $\endgroup$ – Dmitri Pavlov Feb 2 '19 at 19:27
  • 2
    $\begingroup$ @DmitriPavlov Seems reasonable. I don't know much about cubical sets, but the nlab page suggests it's not known if the cubical analog of $Ex^\infty$ is a fibrant replacement, so it might not be known if homotopy classes can be represented this way in cubical sets. $\endgroup$ – Tim Campion Feb 2 '19 at 19:37

I've worked out a proof that a form of cubical $Ex^\infty$ functor is a fibrant replacement functor for cubical sets with connections. From this, it's basically immediate that a homotopy class can be represented as above. I hope to post this to the arxiv soon.

  • $\begingroup$ Er -- rather, this works with the category of cubical sets with connections and extensions (which allow to permute the different dimensions of a cube), which is sufficient to understand homotopy groups of nerves of categories, if not of general simplicial sets. And I should say there's an additional ingredient one needs -- the fact that the cubical object in $Cat$ given by $\Box^n \mapsto [1]^n$ induces the right functor homotopically. $\endgroup$ – Tim Campion Feb 13 '19 at 8:04

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